Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

double d[3333][3333];
int main() {
	int n;
	cin >> n;
	d[0][0] = 1;
	for (int i = 0; i < n; i++) {
		double p;
		cin >> p;
		d[i + 1][0] = (1 - p)*d[i][0];
		for (int j = 1; j < i + 2; j++)
			d[i + 1][j] = p * d[i][j - 1] + (1 - p)*d[i][j];
	}
	auto r = 0.;
	for (int i = n / 2 + 1; i < n + 1; i++)
		r += d[n][i];
	cout.precision(18);
	cout << r << endl;
	return 0;
}